Equality of Integrals implies equality of measures

Question

I have some trouble solving the following problem:
Let $\mu, \nu$ be two measures on the Borel-$\sigma$-algebra on $\mathbb{R}$. Assume that $\mu(\mathbb{R}) = 1$. Prove that, if for all continuous, compactly supported functions $f$ on $\mathbb{R}$, $\int_{\mathbb{R}} f d\mu = \int_{\mathbb{R}} f d\nu$, then $\mu = \nu$.

It is a problem in a series of questions, where I have already proven that a similar version holds: If the quality holds for all functions $f$ such that $f$ is an indicator function on an open interval, then the measures are the same. For this I used the Dynkin Lemma. It's coming from a course about the Basics of Probability Theory. We did measure theory basics and just started with integration/Expectation of random variables, just to give you an overview of what knowledge our prof assumes we have.

My problem lies with continuity. Normally I would try to do something with simple functions, but they aren't continuous on the whole of $\mathbb{R}$. This is my first course on measure theory and I am completely stumped. Any help would be greatly appreciated.

Answer 1

As the comments point out, you can approximate any indicator function with continuous functions. The idea is that you just linearly interpolate by the endpoints. More specifically, to approximate $1_{(a,b)}$, we can define \begin{align*} f_n(x) := \begin{cases} 0 & x < a-\frac{1}{n} \\ n (x-a) + 1 & a-\frac 1n \le x \le a \\ 1 & a < x < b \\ n (b-x) + 1 & b \le x \le b + \frac 1n \\ 0 & x > x + \frac 1n \end{cases}. \end{align*} The functions $f_n(x)$ converge to $1_{(a,b)}(x)$ pointwise, and are all dominated by $f_1$, so by the dominated convergence theorem we have $\int f_n d\mu \rightarrow \int 1_{(a,b)}d\mu$ and $\int f_n d\nu \rightarrow \int 1_{(a,b)}d\nu$. Since each $f_n$ is continuous, this shows $\int 1_{(a,b)}d\mu = \int 1_{(a,b)}d\nu$.

Answer 2

This holds even in more general setting. Suppose $\mu$ is a regular Borel measures on a locally compact Hausdorff space $X$, then for any open set $U$, we have $$\mu(U)=\sup\left\{\int fd\mu~:~f\in C_c(X),~0\leq f\prec \chi_U\right\}$$ where $f\prec \chi_U$ means $supp(f)\subset U$. This is quite easy to prove. The inequality $\mu(U)\geq\sup\left\{\int fd\mu~:~f\in C_c(X),~0\leq f\prec \chi_U\right\}$ is obivious. To prove the reverse inequality, let $0\leq\alpha<\mu(U)$ be any arbitrary number. Use the regularity of $\mu$ to chose a compact set $K$ such that $\mu(K)>\alpha$. Then, using the local compactness of $X$, choose $g\in C_c(X)$ such that $\chi_K\leq g\prec \chi_U$. Then, $\int gd\mu>\alpha$. Hence, we conclude that $\mu(U)\leq\sup\left\{\int fd\mu~:~f\in C_c(X),~0\leq f\prec \chi_U\right\}$ also holds true, thus proving the assertion.

Now, in your case $\mu,\nu$ are Borel measures which are finite on compact set, hence they are regular. Now using the previous assertion, we know that $\mu(U)=\nu(U)$ for all open sets and hence $\mu=\nu$ using the regularity.